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bpo-47053: Refactor BINARY_OP_INPLACE_ADD_UNICODE (GH-32122)

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Dennis Sweeney 2022-03-28 22:07:05 -04:00 • committed by GitHub
parent bad86a621a
commit 788154919c
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1 changed files with 18 additions and 14 deletions
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32
Python/ceval.c
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@ -2003,28 +2003,32 @@ handle_eval_breaker:
DEOPT_IF(!PyUnicode_CheckExact(left), BINARY_OP);
DEOPT_IF(Py_TYPE(right) != Py_TYPE(left), BINARY_OP);
_Py_CODEUNIT true_next = next_instr[INLINE_CACHE_ENTRIES_BINARY_OP];
int next_oparg = _Py_OPARG(true_next);
assert(_Py_OPCODE(true_next) == STORE_FAST ||
_Py_OPCODE(true_next) == STORE_FAST__LOAD_FAST);
/* In the common case, there are 2 references to the value
* stored in 'variable' when the v = v + ... is performed: one
* on the value stack (in 'v') and one still stored in the
* 'variable'. We try to delete the variable now to reduce
* the refcnt to 1.
*/
PyObject *var = GETLOCAL(next_oparg);
DEOPT_IF(var != left, BINARY_OP);
PyObject **target_local = &GETLOCAL(_Py_OPARG(true_next));
DEOPT_IF(*target_local != left, BINARY_OP);
STAT_INC(BINARY_OP, hit);
GETLOCAL(next_oparg) = NULL;
/* Handle `left = left + right` or `left += right` for str.
*
* When possible, extend `left` in place rather than
* allocating a new PyUnicodeObject. This attempts to avoid
* quadratic behavior when one neglects to use str.join().
*
* If `left` has only two references remaining (one from
* the stack, one in the locals), DECREFing `left` leaves
* only the locals reference, so PyUnicode_Append knows
* that the string is safe to mutate.
*/
assert(Py_REFCNT(left) >= 2);
Py_DECREF(left); // XXX never need to dealloc
STACK_SHRINK(1);
PyUnicode_Append(&TOP(), right);
STACK_SHRINK(2);
PyUnicode_Append(target_local, right);
Py_DECREF(right);
if (TOP() == NULL) {
if (*target_local == NULL) {
goto error;
}
JUMPBY(INLINE_CACHE_ENTRIES_BINARY_OP);
// The STORE_FAST is already done.
JUMPBY(INLINE_CACHE_ENTRIES_BINARY_OP + 1);
NOTRACE_DISPATCH();
}